Question

Prove that the squares of all the terms of the arithmetic sequence 4,7,10,.... belong to the sequence​

Answer 1

Answer:

a = first term = 4

d = common difference = 7-4 = 3

an = a + (n-1) × d

an = 4 + 3n-3 = 3n + 1

(an)2 = (3n + 1)2 = 9n2 + 6n + 1 = 3n(3n + 2) + 1 = 3{n(3n + 2)} + 1

As, (an)2 is also of the form 3n + 1, the squares of all the terms of the arithmetic sequence 4, 7, 10, … belong to the sequence.

Answer 2

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$\huge\fcolorbox{pink}{Red}{ ★Q᭄uestion✍︎}$

Prove that the squares of all the terms of the arithmetic sequence 4,7,10,.... belong to the sequence

$\huge\fcolorbox{pink}{Red}{ ★A᭄ΠSWΣR✍︎}$

AP is

4,7,10 ....

$a = 4 d = 3$

$nth \: term = a +(n-1)d = 4 + (n-1)3 = 4 + 3n - 3 = \: 3n+1$

$square \: of \: nth \: Term = (3n + 1)²$

$= 9n² + 6n +1 \\ </p><p></p><p>= 3(3n² + 2n) + 1$

$3n² + 2n = k</p><p></p><p>= 3k + 1$

$square \: of \: nth \: Term = kth \: Term</p><p></p><p>+ 2n$

$where \: k = 3n²$

Hence:- ,

proved that the squares of all the terms of the arithmetic sequence 4,7,10....belong to the sequence

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$\fcolorbox{pink}{green}{ ★AnSWΣRED By @Missfairy01 ☆}$